Solve for $x$ and $y$ using substitution. ${-x+5y = 1}$ ${x = y+3}$
Since $x$ has already been solved for, substitute $y+3$ for $x$ in the first equation. ${-}{(y+3)}{+ 5y = 1}$ Simplify and solve for $y$ $-y-3 + 5y = 1$ $4y-3 = 1$ $4y-3{+3} = 1{+3}$ $4y = 4$ $\dfrac{4y}{{4}} = \dfrac{4}{{4}}$ ${y = 1}$ Now that you know ${y = 1}$ , plug it back into $\thinspace {x = y+3}\thinspace$ to find $x$ ${x = }{(1)}{ + 3}$ ${x = 4}$ You can also plug ${y = 1}$ into $\thinspace {-x+5y = 1}\thinspace$ and get the same answer for $x$ : ${-x + 5}{(1)}{= 1}$ ${x = 4}$